Finding and area given (integration in polar coordinates) There is an area given:
$$S = \big\{ (x,y):(x^2+y^2)^2 \le 2(x^2-y^2)\big\}$$
I know that I have to use polar coordinate system and I know that:
$$r \le\sqrt{2\cos2\phi}$$
However I don't know how to find the range of $\phi$.
Is there a method to draw that kind of areas?
 A: You have
$$
P:\{(r,\phi)\in[0,\infty)\times[0,2\pi]~:~r^2\leq 2\cos(2\phi)\}\to S,~P(r,\phi)=(r\cos(\phi),r\sin(\phi)).
$$
So you have to restrict your domain to get $S$ as the image of $P$. It looks like here:

To get the range for $\phi$ you have to consider $r^2\leq 2\cos(2\phi)$. The LSH is positive, so the inequality can just hold if and only if $\cos(2\phi)\geq 0$. Since $\phi\in[0,2\pi]$ you get $2\phi\in[0,4\pi]$ and 
$$
\cos(2\phi)\geq 0 \Leftrightarrow 2\phi\in\left[0,\frac12\pi\right]\cup\left[\frac32\pi,\frac52\pi\right]\cup\left[\frac72\pi,4\pi\right]\\
\Leftrightarrow \phi\in\left[0,\frac14\pi\right]\cup\left[\frac34\pi,\frac54\pi\right]\cup\left[\frac74\pi,2\pi\right].
$$
So you get
$$
M:=\{(r,\phi)\in[0,\infty)\times[0,2\pi]~:~r^2\leq 2\cos(2\phi)\}\\
=\left\{(r,\phi)\in[0,\infty)\times\left(\left[0,\frac14\pi\right]\cup\left[\frac34\pi,\frac54\pi\right]\cup\left[\frac74\pi,\frac2\pi\right]\right)~:~r^2\leq 2\cos(2\phi)\right\}.
$$
Now you get
$$
\int_S1~d(x,y)=\int_M r~d(r,\phi)\\
=\int_0^{\frac14\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi+\int_{\frac34\pi}^{\frac54\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi+\int_{\frac74\pi}^{2\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi
$$
Because of the symmetry you can simplify
$$
\int_S1~d(x,y)=4\int_0^{\frac14\pi}\int_0^\sqrt{2\cos(2\phi)}r~dr~d\phi.
$$
Result:

 The area should be $2$.

